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For $s,\tau \ge 0$, use conditional probability and the survival function:It is independent of $s$ and equals $P(T>\tau )$. This is the memoryless property of the exponential distribution, equivalent here to a constant hazard function $\alpha$.

Write $\lambda =\ln ({\rho }^{-1})>0$. The probability density function normalizes when $1=B{\int }_0^{\infty }{e}^{-\lambda x}dx=B/\lambda$, so $B=\lambda$. Its survival function is $e^{-\lambda x}$ for $x\ge 0$; hence, for $x>y\ge 0$,Differentiating this conditional tail givesThe value assigned at the single endpoint does not affect any probability. The memorylessness of the exponential distribution says that $X-y$, conditional on $X>y$, is again an exponential distribution of rate $\lambda$. Its first two moments are $1/\lambda$ and $2/{\lambda }^2$, found by integration by parts, so translation leaves the variance unchanged:

Write $\bar{s}>0$ for the mean in minutes. An exponential distribution with that mean has rate $1/\bar{s}$, so its probability density function isThe density has units of inverse minutes, ensuring that its integral is a dimensionless probability.

The first arrival time $T_1$, measured in hours after opening, has exponential distribution of rate twelve. ThereforeEquivalently, there must be no arrivals in the first quarter hour and at least one in the next; independent increments give $e^{-3}(1-e^{-3})$.

Put $Y_n=n(1-M_n)\ge 0$. For fixed $x\ge 0$ and sufficiently large $n>x$,For $x<0$ its distribution function is zero. The limiting distribution function is therefore $F(x)=0$ for $x<0$, $F(x)=1-e^{-x}$ for $x\ge 0$, the exponential distribution of rate one. Since this distribution function is continuous everywhere,

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}[e^{tX}]$ wherever this expectation is finite. Existence on an open interval containing zero is the condition needed for the usual uniqueness and convergence theorems. For the exponential distribution with rate $\lambda >0$,For $t\ge \lambda$ the integral diverges. Differentiating at zero gives mean $1/\lambda$ and variance $1/{\lambda }^2$.

The independent identically distributed central limit theorem states that for variables of common finite mean $\mu$ and finite positive variance ${\sigma }^2$,For the exponential distribution, $\mu =1/\lambda$ and $\sigma =1/\lambda$. Thus the event in question is exactlyThe limiting normal distribution has a continuous distribution function at every real $x$, so convergence in distribution givesHere $\Phi (x)=(2\pi {)}^{-1/2}{\int }_{-\infty }^x{e}^{-u^2/2}du$, including the strict-inequality convention because the limiting distribution is continuous.

The interarrival times are independent and identically distributed with exponential distribution of rate $\lambda$: for any finite $k$ their joint density is ${\lambda }^k\exp (-\lambda {\sum }_{j=1}^k{x}_j)$ on $x_j>0$. In particular

Use independent colour marks, also independent of arrival times, as in the standard marked-process model. Let $K$ be the index of the first white arrival. Then $\mathbb{P}(K=k)=p(1-p{)}^{k-1}$ and $K$ is independent of the interarrival times. Thus $T={\sum }_{j=1}^K{X}_j$. Conditional on $K$, independence makes its moment-generating function a power of $M_X$, and summing the geometric series givesThis is the exponential distribution of rate $p\lambda$, so $T\sim \mathrm{Exp}(p\lambda )$. The calculation assumes $p>0$; if $p=0$, there is no white arrival and $T=+\infty$ almost surely.

The deductive-nomological model represents an explanation as a valid deduction of an event description from scientific laws and particular circumstances. In the schemathe premises are the explanans and $E$ the explanandum. At least one law must be essential to the derivation, the premises must have empirical content, and a correct explanation must have true premises. Carl Hempel and Paul Oppenheim distinguish these adequacy requirements from merely giving a valid argument. Their developed 1948 analysis concerns deductive explanation; its name should not be extended without qualification to every statistical explanation.^[ref]

The model captures something valuable. For an ideal simple harmonic oscillator, the equation of motion $m\ddot{x}=-kx$, a positive mass $m$, a positive spring constant $k$ and non-equilibrium initial conditions determine a periodThe derivation explains why the period changes with those parameters within the model. It is not just the report that an oscillation occurred. Since a real spring may have damping or nonlinear response, applying the explanation requires justification of the approximation. The model's truth condition cannot turn an idealized calculation into a literally exact account of every laboratory spring.

The account is not sufficient to settle explanatory direction. For a vertical pole of height $h$ on level ground, a solar elevation angle $0<\theta <\pi /2$ and parallel light rays give a shadow length $\ell =h\cot \theta$. Given $h$ and $\theta$, one can deduce $\ell$; given $\ell$ and $\theta$, one can equally deduce $h$. Both calculations can use the same true optical and geometrical assumptions. Yet normally the pole and illumination explain the shadow, whereas measuring the shadow provides evidence about the pole's height. Deducibility does not distinguish this evidential inference from a causal explanation. This is the problem of explanatory asymmetry.

Nor is strict deduction necessary for every good physical explanation. A radioactive decay model can explain a sample's decay rate and the chance of a nucleus decaying in a specified interval, while not entailing which particular nucleus decays first. The exponential distribution supplies a controlled statistical account. Adding the observed result as a premise would permit a deduction but would not supply the missing explanation. An inductive-statistical model is a different proposal, and even that needs an account of which statistical information is relevant.

These objections do not show that every explanation must instead recount a sequence of causes. Explanations based on conservation laws or symmetry can constrain a phenomenon without specifying a detailed mechanism. The important point is that the role of the premises, the question being asked and the relevant dependence matter, in addition to logical consequence. The deductive-nomological model is a useful pattern for some physical explanations, but it is not an adequate necessary-and-sufficient account of explanation throughout the physical sciences.

The laboratory is the parent's rest frame. In this relativistic two-body decay, conservation of momentum gives equal and opposite daughter momenta, of magnitude $p$. Their total relativistic energies satisfySubtracting the relativistic energy-momentum relations eliminates $p$:Solving the two linear equations givesSince three-momentum obeys $p{c}^2=Ev$, the daughter speeds areTheir directions are opposite; the initial state does not specify their common axis. The kinetic energies are $K_1=5m{c}^2/8$ and $K_2=3m{c}^2/8$. Their sum $m{c}^2$ is the loss of rest energy when the total daughter rest mass is $3m$ instead of $4m$.

The proper mean lifetime ${\tau }_0$ is the mean lifetime measured in the unstable particle's own rest frame, or equivalently the mean elapsed proper time before decay. In the exponential decay model, its survival probability to proper time $\tau$ is $e^{-\tau /{\tau }_0}$. For the daughter of rest mass $m$, the Lorentz factor is ${\gamma }_1=E_1/(m{c}^2)=13/8$, so time dilation gives a laboratory mean lifetime of ${\gamma }_1{\tau }_0=3.25\mu \mathrm{s}$.

At constant speed $v_1$, travelling a distance $L$ requires laboratory time $L/v_1$ and proper time $L/({\gamma }_1{v}_1)$. Its mean decay length isfor ${\tau }_0=2\times 10^{-6}\mathrm{s}$. HenceThe mean lifetime sets the scale of an exponential distribution; it is not a fixed deadline, so travelling beyond one mean decay length has a nonzero probability.

An exponential distribution of rate $\lambda$ is a Gamma distribution with shape one. Repeated additivity of independent gamma distributions with a common rate therefore givesWith integer shape this is also called an Erlang distribution.

The Poisson interarrival times of a homogeneous Poisson process are independent and identically distributed with the exponential distribution of rate $\lambda$. Consequently $Y_1,Y_2$ are independent random variables and their joint probability density isHere $Y_2$ is the second interarrival duration, rather than the time of the second arrival.

Put $Z_n=n(1-M_n)$. Its survival function isFor each fixed $z\ge 0$, eventually $n\ge z$, and the exponential limit gives $(1-z/n{)}^n\to e^{-z}$. Thus its cumulative distribution function tends to $0$ for $z<0$ and to $1-e^{-z}$ for $z\ge 0$. This continuous limiting cumulative distribution function identifies the exponential distribution with rate one:This is the exponential endpoint limit of a uniform maximum: the gap from the upper endpoint is typically of order $1/n$, although the maximum itself approaches a constant.

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.

By independence of random variables, the joint probability density is the product of the two probability density functions of the rate-one exponential distributions:Its values on the boundary are immaterial, because the boundary has zero area.

Every $X_n$ has the same exponential distribution, so $X_n\stackrel{d}{\to }X$ for $X\sim \mathrm{Exp}(1)$. Part (ii) gives $Z_n-X_n\stackrel{p}{\to }0$. Applying Slutsky's theorem to $Z_n=X_n+(Z_n-X_n)$ yieldsThe limiting probability distribution does not depend on $c$. This is precisely the attraction to the exponential invariant law of a random affine recursion.

The update defines a time-homogeneous Markov process: given its present value, the next value depends only on independent fresh input. Part (b)(i) identifies an invariant law: initializing with the rate-one exponential distribution preserves that distribution at every time. Part (b)(iii) shows convergence of the one-time laws to this invariant law from every deterministic initial value.

For a countable state space, the analogous countable-state Markov chain convergence theorem says that an irreducible Markov chain that is aperiodic and positive recurrent converges to its unique stationary distribution from every initial state. Here the state space is continuous, so that countable-state theorem does not directly supply the proof; the explicit coupling does. The two conclusions have the same equilibrium interpretation.

Set $T_0=0$ and $E_j=T_j-T_{j-1}$. The Poisson interarrival times $E_j$ are independent and identically distributed random variables with the rate-one exponential distribution. In particular, $\mathbb{E}{E}_j=1$ and $\mathrm{Var}(E_j)=1$, so the finite, positive variance hypothesis of the central limit theorem holds. Since $T_n={\sum }_{j=1}^n{E}_j$,

The support condition $-y<x<y$ forces $y>0$. As a check, the joint probability density integrates toFor any real $x$, the marginal density of $X$ isDividing the joint probability density by this marginal density gives the conditional density:Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two exponential distribution.

For $U=X^2$, use the two inverse branches $x=\sqrt{u}$ and $x=-\sqrt{u}$. Each branch has absolute Jacobian determinant $1/(2\sqrt{u})$. Both lie in the original support exactly when $u>0$ and $y>\sqrt{u}$. The change of variables for probability densities therefore yieldsNo atom of a measure is created at $u=0$, since $\mathbb{P}(X=0)=0$. The value of the probability density function at this boundary may be assigned arbitrarily. Checking the integral over $0<u<y^2$ gives $4y{e}^{-2y}$, which integrates to one.